Question 1

Azeotropy and Raoult’s law: a binary azeotrope that boils at a temperature lower than either pure component exhibits which type of deviation from Raoult’s law?

Accepted Answer

Introduction / Context: Azeotropes are constant-boiling mixtures that behave non-ideally relative to Raoult’s law. Whether an azeotrope boils lower or higher than its pure components depends on the sign of deviation from ideal solution behavior, which is crucial for separation strategy decisions. Given Data / Assumptions: Binary mixture forms an azeotrope. Azeotropic boiling point is lower than that of either pure component at the same pressure. Raoult’s law describes ideal solutions; deviations reflect intermolecular interaction changes on mixing. Concept / Approach: Positive deviation means that component–component interactions are weaker than like–like interactions; the total vapor pressure is higher than predicted by Raoult’s law, leading to a lower boiling temperature. Hence, a minimum-boiling azeotrope corresponds to positive deviation. Negative deviation yields a maximum-boiling azeotrope (higher boiling than either component). Step-by-Step Solution: Identify azeotrope type: minimum-boiling (lower than both components).Relate to deviation: minimum-boiling ↔ positive deviation.Reason: weaker interactions increase volatility (higher total pressure at a given T).Therefore, select positive deviation. Verification / Alternative check: Phase diagrams for ethanol–water (minimum-boiling) exemplify positive deviation from Raoult’s law. Why Other Options Are Wrong: Negative deviation corresponds to maximum-boiling azeotropes. No deviation is the ideal case, which does not form azeotropes. “Cannot be predicted” is incorrect given the information. Common Pitfalls: Confusing minimum- and maximum-boiling azeotropes; assuming all azeotropes have the same deviation sign. Final Answer: Positive deviation

Question 2

Mixing tank internals: as a preliminary design rule for baffled cylindrical vessels, the total baffle width is typically selected as what fraction of the tank diameter?

Accepted Answer

Introduction / Context: Baffles suppress vortex formation and convert swirling into efficient top-to-bottom circulation in stirred tanks. Correct baffle sizing is essential to achieve target mixing times and reliable scale-up. A common rule-of-thumb specifies baffle width as a fraction of tank diameter. Given Data / Assumptions: Baffled, cylindrical stirred tank with four flat baffles. General-purpose mixing (not special viscous or gas–liquid exceptions). Looking for standard preliminary proportioning. Concept / Approach: Standard practice sets individual baffle width at roughly 0.1–0.12 of the tank diameter. This provides strong flow disruption without excessive blockage or power draw. Smaller baffles may permit vortexing; much larger baffles can increase power requirements and hinder flow around the tank periphery. Step-by-Step Solution: Define d = tank diameter.Choose baffle width w ≈ 0.1 d to 0.12 d.Verify against impeller type and Reynolds number; adjust slightly if needed.Accept 0.1–0.12 as the default design range. Verification / Alternative check: Vendor guidelines and mixing handbooks align with w/d ≈ 0.1–0.12 for standard turbulent mixing. Why Other Options Are Wrong: 0.2–0.45 or higher values are unusually large and increase blockage and power. Very small values (0.02–0.05) are ineffective at suppressing swirl. Common Pitfalls: Using the same baffle width for highly viscous systems; not accounting for nozzle intrusions or internals that already impede swirl. Final Answer: 0.1–0.12

Question 3

Hydraulic diameter concept: for steady flow through a duct of constant noncircular cross-section with flow area A and wetted perimeter P, which expression defines the equivalent (hydraulic) diameter?

Accepted Answer

Introduction / Context: The hydraulic (equivalent) diameter allows engineers to apply circular-tube correlations to noncircular ducts by mapping geometry via area and wetted perimeter. This is widely used for heat transfer and pressure drop calculations in channels, manifolds, and HVAC ducts. Given Data / Assumptions: Fully wetted perimeter P with internal flow. Constant cross-section with flow area A. Turbulent/laminar correlations typically expressed in terms of Re and Nu using Dh. Concept / Approach: The standard definition is Dh = 4 A / P. This preserves key dimensionless groups and reduces to the tube inside diameter for a circular tube (A = πD^2/4, P = πD → Dh = D). Using any other ratio would distort Reynolds number and friction factor predictions for noncircular ducts. Step-by-Step Solution: State definition: Dh = 4 A / P.Check circular tube limit → recovers D exactly.Apply Dh consistently in Re = ρ v Dh / μ and in heat-transfer correlations.Hence, choose 4 A/P. Verification / Alternative check: Standard fluid mechanics and heat transfer texts present Dh = 4A/P as the universal definition for internal flows. Why Other Options Are Wrong: A/P or P/A change dimensions improperly. 4 P/A is inverse of the correct form. A alone lacks perimeter information and is not a length scale. Common Pitfalls: Using external perimeter instead of wetted perimeter; misapplying Dh to partially filled conduits where free surface exists. Final Answer: 4 A/P

Question 4

Effect of increasing liquid depth on distillation trays: what is the principal hydraulic consequence of having a larger liquid head on trays in an atmospheric column?

Accepted Answer

Introduction / Context: Tray hydraulics govern pressure drop, capacity, and mass-transfer efficiency in distillation. The liquid head (depth) on a tray directly impacts the vapor head that must be overcome, influencing the allowable throughput and energy usage. Understanding this relationship is fundamental for troubleshooting and debottlenecking. Given Data / Assumptions: Atmospheric column with conventional trays (sieve, valve, or bubble-cap). Comparing conditions with relatively larger versus smaller liquid depth on the tray. Other variables (weir length, froth behavior) being equal for the conceptual discussion. Concept / Approach: Increasing liquid head increases the static and dynamic heads that vapor must overcome to pass through tray perforations or caps, thereby increasing pressure drop per tray. While some liquid depth is needed to avoid weeping and to stabilize froth, excessive depth generally increases hydraulic losses and may reduce stage efficiency due to backmixing and longer liquid residence causing dispersion. Step-by-Step Solution: Recognize that ΔP_tray ≈ dry ΔP + aerated liquid head components.Higher liquid depth → higher aerated head → higher ΔP.Efficiency trend is not necessarily improved by deep liquid; too deep often hurts efficiency.Therefore, the robust statement is that pressure drop increases. Verification / Alternative check: Tray design correlations (e.g., Fair, Bennett–Myers) reflect increasing ΔP with increasing clear liquid height and froth density. Why Other Options Are Wrong: “Leads to high tray efficiency” is not generally valid; deep liquid may lower efficiency. “Both” is incorrect for the reason above. “Neither” contradicts hydraulic fundamentals. “Eliminates entrainment completely” is false; entrainment depends on vapor velocity and froth properties. Common Pitfalls: Assuming more liquid always improves mass transfer; ignoring the trade-off between weeping control and hydraulic losses. Final Answer: Results in higher pressure drop per tray

Question 5

Typical pressure drop per tray: in an atmospheric distillation column, what is the approximate pressure drop across a single tray, expressed in psi?

Accepted Answer

Introduction / Context: Knowing the typical pressure drop per tray helps in estimating overall column pressure profile, condenser duty, and potential vacuum limitations. Atmospheric columns are designed to keep per-tray pressure drops modest to prevent excessive cumulative losses over many stages. Given Data / Assumptions: Atmospheric operation (near 1 atm at the top). Conventional tray types and normal vapor/liquid loadings. Pressure drop requested in psi per tray. Concept / Approach: Empirical design practice places per-tray pressure drop in a relatively small band. For many atmospheric services, pressure drop of roughly 0.07–0.12 psi per tray is typical, balancing sufficient vapor–liquid contact with reasonable column height and energy use. Much higher drops are more typical of high-pressure or high-capacity services and would be undesirable at atmospheric conditions. Step-by-Step Solution: Identify normal ΔP range for atmospheric tray service.Select the modest band 0.07–0.12 psi consistent with standard practice.Exclude larger values (≥ 0.5 psi) representing unusually high load or different service.Confirm selection. Verification / Alternative check: Rules-of-thumb in distillation design texts list comparable ranges for atmospheric columns. Why Other Options Are Wrong: 0.01–0.5 is too broad; 0.01 psi is unrealistically low and 0.5 psi is too high for typical atmospheric operation. 0.5–1.0 and 1–3 psi are characteristic of much heavier services and would cause undue total pressure loss. 3–5 is far above normal. Common Pitfalls: Confusing mm H2O or inches of water with psi; not accounting for tray type and loading effects. Final Answer: 0.07 – 0.12

Question 6

Shell-and-tube heat exchanger internals: the center-to-center spacing between successive baffles is called the baffle pitch (baffle spacing). Which of the following is NOT a function of baffles?

Accepted Answer

Introduction / Context: Baffles are key internals in shell-and-tube heat exchangers. Their geometry and spacing (baffle pitch) strongly influence shell-side flow pattern, pressure drop, heat transfer, and mechanical stability. Knowing what baffles do—and do not do—prevents design misunderstandings. Given Data / Assumptions: Baffles mounted on the shell side (segmental, double-segmental, or various designs). Tube bundle supported by baffles and tie rods. Tube-side flow is confined within tubes and unaffected directly by baffles. Concept / Approach: Baffles primarily act on the shell-side fluid: they create crossflow over tubes, increase shell-side turbulence and heat-transfer coefficient, control leakage and bypass streams, and support the tubes to mitigate vibration. They do not act directly on the tube-side flow; tube-side coefficients are governed by tube flow velocity, tube roughness, and fluid properties, not by shell baffles. Step-by-Step Solution: Identify shell-side roles: flow redirection, turbulence, and tube support.Recognize mechanical benefits: baffles restrain tube spans, lowering vibration risk.Note separation of circuits: baffles cannot influence tube-side turbulence directly.Hence, the non-function is “increase tube-side coefficient.” Verification / Alternative check: TEMA and exchanger design references list baffle objectives specifically for shell-side performance and structural support. Why Other Options Are Wrong: Residence time increase on shell side: a consequence of longer flow path via crossflow/zig-zag. Support to tube bundle: a primary mechanical function. Vibration reduction: achieved by limiting tube unsupported span and altering flow-induced excitation. Directing crossflow on shell side: central to enhancing shell-side heat transfer. Common Pitfalls: Assuming baffles affect tube-side performance; overlooking pressure-drop penalties from too-tight baffle spacing. Final Answer: To increase the tube-side heat-transfer coefficient by inducing turbulence

Question 7

In a shell-and-tube heat exchanger, if a single tube pass is converted to a two-pass arrangement (same total tube length and same overall flow rate), by what factor will the tube-side pressure drop per unit length increase?

Accepted Answer

Introduction / Context: In heat exchanger design, changing the number of tube passes affects flow distribution, velocity, and frictional losses. This question tests the understanding of how converting from a single-pass to a two-pass tube arrangement impacts tube-side pressure drop for the same overall flow rate and tube bundle length. Given Data / Assumptions: Total volumetric flow rate on tube side remains the same. Total tube length in the flow path is effectively traversed twice in a two-pass exchanger. Number of tubes per pass halves in a two-pass arrangement, so flow per tube doubles. Flow is turbulent so friction factor is treated approximately constant for a first-cut estimate; pressure drop scales with v^2. Concept / Approach: For fully developed turbulent flow in a smooth tube, pressure drop per tube can be approximated as: ΔP ∝ (L/D) * (ρ * v^2 / 2). Converting to two passes: (1) Path length along a tube doubles (L → 2L), and (2) Velocity in each tube doubles because only half as many tubes are in parallel (v → 2v). Step-by-Step Solution: Single-pass baseline: ΔP₁ ∝ L * v^2.Two-pass length effect: L → 2L gives a factor of 2.Two-pass velocity effect: v → 2v gives v^2 → (2v)^2 = 4v^2.Combined factor: 2 (length) * 4 (velocity-squared) = 8.Therefore, ΔP₂ ≈ 8 * ΔP₁ for the same total flow rate. Verification / Alternative check: A quick network analogy: halving parallel paths doubles flow per path (velocity roughly doubles). Two series segments (two passes) double the frictional contribution. Multiplying effects again gives 8×, confirming the result. Why Other Options Are Wrong: Increase by 4: ignores the doubled path length.Increase by 2: considers only the added pass length, not velocity change.Approximately unchanged: contradicts stronger velocity and length effects.Decrease by 2: physically implausible for the stated conversion. Common Pitfalls: Forgetting that tubes per pass are halved in two-pass designs.Assuming friction factor changes cancel the velocity effect; they generally do not for a first estimate.Comparing total exchanger pressure drop but not normalizing per unit length consistently. Final Answer: Increase by 8

Question 8

In the downcomer of a distillation column, the minimum recommended liquid residence time is approximately how many seconds for stable operation and proper disengagement?

Accepted Answer

Introduction / Context: Downcomers in tray columns carry liquid from one tray to the next. Adequate residence time in the downcomer is essential for hydraulic stability, foam collapse, and gas–liquid disengagement. Designers use a minimum recommended residence time to avoid backup and weeping/flooding issues. Given Data / Assumptions: Conventional sieve or valve tray operation. Ambient-to-moderate pressure with typical hydrocarbon or aqueous systems. Rule-of-thumb design criteria from standard tray hydraulic practices. Concept / Approach: Empirical guidelines suggest a minimum downcomer residence time around a few seconds to allow hydrodynamic stabilization. Too short a residence time raises downcomer exit turbulence and can lead to jet entrainment onto the next tray, while too long wastes area. Step-by-Step Solution: Select a commonly accepted minimum residence time for typical tray columns.Industry practice often targets roughly 3–5 s, with 5 s used as a conservative minimum for design.Hence the recommended value is about 5 seconds. Verification / Alternative check: Tray hydraulics correlations (e.g., capacity and weeping/flooding checks) are consistent with several seconds of residence for stable downcomer operation, especially at turndown conditions. Why Other Options Are Wrong: 2.5: Often too short; increases risk of instability and entrainment.7.5 / 12.5 / 17.5: Unnecessarily large; wastes cross-sectional area and can reduce tray active area. Common Pitfalls: Treating residence time independently of froth height and downcomer backup.Ignoring turndown—columns need stability over a range of rates.Using a single value without cross-checking with backup height and capacity limits. Final Answer: 5

Question 9

A bubble-cap tray column is best described as which type of mass-transfer contactor?

Accepted Answer

Introduction / Context: Tray columns—such as sieve, valve, and bubble-cap trays—are widely used in distillation and absorption. Understanding whether they represent finite equilibrium stages or differential (continuous) contacting devices is fundamental to column modeling and design. Given Data / Assumptions: Conventional tray operation at steady state. Each tray provides a discrete contacting zone with vapor bubbling through liquid. Stage efficiency may be below 100%, but the modeling framework treats each tray as an equilibrium stage with an efficiency factor. Concept / Approach: Tray columns are modeled using the equilibrium-stage (or Murphree efficiency) approach, where each tray approximates a finite stage. In contrast, packed columns are modeled as differential contactors using height-of-a-transfer-unit (HTU) and number-of-transfer-units (NTU) concepts. Step-by-Step Solution: Identify device type: Bubble-cap tray has discrete stages (trays).Modeling method: Equilibrium-stage with tray efficiency, not differential balance along height.Hence, the correct description is a finite stage contactor. Verification / Alternative check: Design simulators (stage-based) natively represent tray columns as finite stages, while packed columns rely on rate-based or HTU/NTU frameworks—confirming the classification. Why Other Options Are Wrong: Used only for distillation: false; trays are also used in absorption/stripping.Differential stage contactor: describes packed beds, not trays.Continuous contactor: again, characteristic of packed columns.Packed-bed equivalent: operation differs fundamentally in hydraulics and modeling. Common Pitfalls: Confusing practical inefficiency with the modeling paradigm—efficiency modifies, not overturns, stage-based modeling.Assuming exclusive service (distillation only); trays are versatile. Final Answer: A finite stage contactor

Question 10

For 1.5 inch shell-and-tube heat exchanger tubes, how does the inside flow area change as BWG (Birmingham Wire Gauge) decreases (i.e., wall gets thicker)?

Accepted Answer

Introduction / Context: Tube wall thickness is commonly specified by BWG. A lower BWG number corresponds to a thicker wall. The internal diameter determines the flow area and therefore velocity and pressure drop for a given volumetric flow rate. Given Data / Assumptions: Nominal outside diameter fixed at 1.5 inch. BWG decreases → wall thickness increases. No corrosion allowance changes considered; focus is purely geometric. Concept / Approach: Inner diameter ID = OD − 2 * t. Increasing the wall thickness t reduces ID linearly. Flow area A = (π/4) * ID^2 therefore decreases with decreasing BWG (thicker wall). Step-by-Step Solution: Start with fixed OD = 1.5 in.Decrease BWG → increase t.ID = OD − 2t decreases as t increases.A ∝ ID^2; hence the inside flow area decreases. Verification / Alternative check: Sample numbers: If t increases by 0.05 in, ID drops by 0.10 in, which reduces area roughly proportionally to the square of ID, confirming the trend. Why Other Options Are Wrong: Increases / Remains the same: contradict geometry.Cannot be determined: basic geometry determines it.Increases only at high Reynolds number: flow regime does not reverse the geometric effect. Common Pitfalls: Confusing BWG numbering (lower number means thicker wall).Mixing up OD and ID when applying thickness changes. Final Answer: Decreases

Question 11

For a turbine agitator installed in a tank of diameter d, the typical impeller diameter range used in design is:

Accepted Answer

Introduction / Context: Turbine agitators are radial-flow impellers commonly used for gas dispersion, liquid blending, and solid suspension. Selecting an appropriate impeller diameter relative to tank diameter ensures efficient power draw and mixing without excessive shear or mechanical penalties. Given Data / Assumptions: Cylindrical tank diameter = d. Standard six-blade Rushton or similar turbine. No unusual baffle or multiple-impeller special cases considered. Concept / Approach: Empirical mixer design practice places turbine impellers in the range of roughly one-third to one-half of tank diameter to balance flow circulation, tip speed, and power number. Much larger impellers approach the tank wall causing inefficient flow; much smaller impellers underutilize the volume. Step-by-Step Solution: Identify accepted design range for turbine impellers.Use rule of thumb: D_impeller ≈ 0.3 d to 0.5 d.Therefore, the correct answer is 0.3 to 0.5 d. Verification / Alternative check: Mixing handbooks and vendor guidelines routinely recommend roughly 0.33 d to 0.5 d for single-stage turbines in baffled tanks. Why Other Options Are Wrong: 0.1–0.2 d: too small; poor circulation and long blend times.0.65–0.85 d or 0.95 d: too large; wall interference and excessive power draw.0.05–0.08 d: far too small for practical mixing. Common Pitfalls: Forgetting to adjust diameter when multiple impellers are installed at different elevations.Ignoring baffles; unbaffled tanks require different considerations. Final Answer: 0.3 to 0.5 d

Question 12

Which tube layout in a shell-and-tube heat exchanger generally provides the highest heat transfer coefficient on the shell side (for a given shell size)?

Accepted Answer

Introduction / Context: Tube layout determines tube count, crossflow geometry, and shell-side turbulence. Designers balance heat transfer, pressure drop, fouling, and cleanability when selecting layouts such as triangular, square, or rotated square pitch. Given Data / Assumptions: Similar tube outside diameter and pitch ratios. Same shell diameter for comparison. Clean service comparison without unusual constraints. Concept / Approach: Triangular pitch packs tubes most densely, increasing velocity and turbulence for a given shell size, raising the shell-side heat transfer coefficient. Square layouts ease mechanical cleaning but have fewer tubes and lower turbulence at the same shell size. Step-by-Step Solution: Compare pack density: triangular > square.Higher tube count at fixed shell diameter drives higher crossflow velocity.Higher velocity and turbulence enhance heat transfer, making triangular pitch superior for heat duty (though with higher pressure drop). Verification / Alternative check: Standards and design texts consistently state triangular pitch provides highest shell-side coefficients but at the expense of cleanability and pressure drop. Why Other Options Are Wrong: Square / rotated square: lower tube density for a given shell size → lower shell-side h.Independent of layout: incorrect; layout affects both h and ΔP.Random pitch: not a standard engineered option. Common Pitfalls: Ignoring fouling/cleaning needs; square pitch may be preferred for mechanical cleaning even with lower h.Equating “highest heat transfer” with “best design” without considering pressure drop limits. Final Answer: Triangular pitch

Question 13

Cylindrical pressure vessels installed horizontally are most commonly supported using which type of support?

Accepted Answer

Introduction / Context: Pressure vessels require supports that safely transmit loads to foundations while accommodating thermal expansion and avoiding excessive local stresses. Horizontal cylindrical vessels use specific supports tuned to shell geometry. Given Data / Assumptions: Horizontal orientation with typical length-to-diameter ratios. Conventional petrochemical service and foundations. No seismic/anchor special designs considered beyond common practices. Concept / Approach: Saddle supports cradle the cylindrical shell over a broad area, distributing loads and limiting localized stresses. One saddle is often fixed (anchored) and the other is sliding to accommodate thermal expansion. Step-by-Step Solution: Identify vessel orientation: horizontal cylindrical.Select standard support type: two saddles spaced near head–shell junctions.Purpose: distribute load and allow expansion (one fixed, one sliding). Verification / Alternative check: Design handbooks specify saddle supports as the default for horizontal drums; skirts are used for vertical vessels; lugs for small equipment; guy wires are not used for pressure vessel dead loads. Why Other Options Are Wrong: Lug: suitable for small vertical vessels—not for large horizontal drums.Skirt: standard for vertical columns, not horizontal shells.Guy wires: not a primary support system for rigid pressure vessels.Ring-only floor grating: not a structural support. Common Pitfalls: Placing saddles too close together, increasing midspan bending.Neglecting sliding saddle to absorb thermal growth, causing nozzle stress. Final Answer: Saddle support

Question 14

Riveted joints: identify the incorrect statement among the following descriptions of joint types and efficiencies.

Accepted Answer

Introduction / Context: Riveted joints are classic permanent fastenings used in boilers, bridges, and structural fabrications. Understanding lap vs. butt joints and the definition of efficiency helps in evaluating joint capability and terminology. Given Data / Assumptions: Lap joint: plates overlap and are riveted directly. Butt joint: plates meet edge-to-edge and are connected with cover straps. Efficiency compares the riveted joint to an unpunched plate of equal width and material. Concept / Approach: Terminology distinguishes joint geometry. A “double riveted lap joint” uses two rows on overlapping plates. A “double riveted butt joint” uses two rows on cover straps over abutting plate edges—not overlap. Step-by-Step Solution: Check each statement against definitions.Riveting is permanent: correct.Overlap with two rows called double riveted butt joint: incorrect—this is a double riveted lap joint.Shank smaller than hole: correct to allow insertion and setting.Efficiency definition: correct. Verification / Alternative check: Standard machine design texts show lap joints as overlapping plates; butt joints employ single or double cover straps with plates meeting edge to edge. Why Other Options Are Wrong: A, C, D, E are aligned with standard definitions; B mislabels a lap joint as a butt joint. Common Pitfalls: Confusing “rows of rivets” with “butt vs. lap” geometry.Ignoring the role of cover plates/straps in butt joints. Final Answer: A joint made by overlapping plates and two rows of rivets is called a double riveted butt joint.

Question 15

Sieve (seive) tray columns are typically designed to operate at what fraction of the flooding vapor velocity to achieve high plate efficiency without excessive entrainment?

Accepted Answer

Introduction / Context: Tray columns have an upper capacity limit known as flooding, where liquid is carried upward by vapor, causing hydraulic instability. Designers select an operating vapor velocity as a percentage of the flooding velocity to balance efficiency, capacity, and pressure drop. Given Data / Assumptions: Conventional sieve trays with typical physical properties and tray spacings. Well-distributed vapor and liquid loads. Focus is on a general design guideline, not a specific system re-rate. Concept / Approach: Operating too close to flooding risks entrainment and downcomer backup; too low reduces mass transfer efficiency due to poor vapor–liquid contact. Practical design targets typically fall in the range of about 60–85% of the predicted flooding velocity for sieve trays, with ~80% a common, satisfactory value. Step-by-Step Solution: Estimate flooding velocity from tray hydraulics correlations.Apply design factor to avoid entrainment and allow turndown margin.Select ~80% of flooding velocity as a robust design point for high efficiency with safety margin. Verification / Alternative check: Vendor and design handbooks frequently recommend 70–85% for sieve trays; choosing 80% lies near the center, offering good performance and margin. Why Other Options Are Wrong: 45%: overly conservative; underutilizes column capacity.60%: workable but leaves more unused capacity than typical designs.95%: too close to flooding; high entrainment and instability risk.30%: far too low, leading to poor contacting. Common Pitfalls: Using a fixed percentage without confirming with tray spacing, weir loading, and downcomer backup.Ignoring physical property changes across the column height. Final Answer: 80 percent of flooding velocity

Question 16

In mechanical design of chemical process vessels, what standard welded joint efficiency is commonly assumed for design calculations (when full radiography is not specified), considering safety and typical fabrication practice?

Accepted Answer

Introduction / Context: Welded joint efficiency is a factor used in the mechanical design of pressure vessels and storage tanks to account for the quality of welds versus base metal. In chemical process equipment design, this factor directly influences required thickness and allowable stress, ensuring safe operation under internal pressure. Given Data / Assumptions: General vessel design context without mandatory full radiography. Standard practice values when only spot or no radiography is specified. Design must remain conservative while reflecting common shop practice. Concept / Approach: The joint efficiency E scales the required thickness: lower E implies thicker walls for the same pressure. Typical values used in design handbooks: E ≈ 1.00 for fully radiographed welds, E ≈ 0.85 for spot radiography or standard shop welds, and lower values when quality control is minimal. Many textbook problems and preliminary designs adopt E = 0.85 as a practical, conservative default when full radiography is not mandated. Step-by-Step Solution: Step 1: Recognize that design thickness t increases as joint efficiency E decreases.Step 2: Identify common reference values: E = 1.00 (full radiography), E ≈ 0.85 (spot radiography/typical), E ≈ 0.7–0.8 (older or less controlled cases).Step 3: For routine chemical equipment design without full radiography, standard practice assumes E ≈ 0.85.Step 4: Select the closest option representing this conventional assumption. Verification / Alternative check: Compare with code tables and typical design examples: preliminary sizing examples often tabulate E = 0.85 when radiography is not specified, increasing required thickness modestly while staying practical for fabrication and cost. Why Other Options Are Wrong: 0.55: Unrealistically low for modern shop welds; would over-penalize thickness.0.75: Possible in some legacy contexts, but not the most common assumed default.0.95: Implies near-full radiography quality; not standard unless explicitly mandated.1.00: Reserved for fully radiographed joints; not the assumed norm here. Common Pitfalls: Using E = 1.00 by default can dangerously under-estimate thickness when radiography is not performed. Conversely, choosing an excessively low E inflates cost and weight without proportional safety benefit. Final Answer: 0.85

Question 17

In gas absorption, for which solute absorbed in water is the overall mass transfer coefficient effectively based only on the gas-film coefficient because the gas-side resistance controls?

Accepted Answer

Introduction / Context: Overall mass transfer during gas absorption depends on resistances in both the gas and liquid films. For highly soluble gases, the liquid-side resistance becomes negligible, leaving the gas-side resistance controlling, so the overall coefficient reflects the gas film coefficient predominantly. Given Data / Assumptions: Absorption into liquid water at typical process conditions. Comparative solubility: NH3 is highly soluble; CO2 and SO2 are moderately soluble with notable liquid-phase equilibria. Film theory: total resistance = gas film + interface + liquid film; dominance depends on solubility and kinetics. Concept / Approach: For very soluble gases, the equilibrium partial pressure at the interface is small, reducing the driving force loss in the liquid phase. Hence, the liquid-film resistance is negligible and the gas-film controls. Ammonia (NH3) dissolves readily in water and reacts to form ammonium hydroxide species, further enhancing apparent solubility and reducing liquid-side resistance. Step-by-Step Solution: Step 1: Rank gases by solubility in water: NH3 >> SO2 ≈ CO2 (context-dependent), H2S varies but less than NH3.Step 2: Identify controlling resistance: highly soluble solutes → gas-film control.Step 3: Conclude that overall K is effectively based on gas-film coefficient for NH3 in water.Step 4: Select NH3 as the correct option. Verification / Alternative check: Design heuristics and classic absorption examples consistently treat NH3–water systems as gas-film controlling in packed towers at moderate conditions, aligning with observed high apparent Henry’s law departure due to chemical interaction. Why Other Options Are Wrong: CO2: Often shows appreciable liquid-film effects and equilibria; not purely gas-film controlled.SO2: Reactive/soluble but commonly not treated as purely gas-film controlled in generic water systems.None of these: Incorrect because NH3 fits.H2S: Soluble, but not typically the standard textbook example for purely gas-film control like NH3. Common Pitfalls: Assuming all acidic gases are gas-film controlled; actual control depends on solubility, reaction, and operating conditions. Always verify with equilibrium and rate data. Final Answer: NH3

Question 18

For flow in non-circular passages, the equivalent diameter defined for pressure-drop calculations is generally __________ the equivalent diameter used for heat-transfer correlations (because the heated perimeter may differ from the wetted perimeter).

Accepted Answer

Introduction / Context: When analyzing non-circular ducts or annuli, engineers use an equivalent (hydraulic) diameter to apply standard pressure-drop and heat-transfer correlations. However, the appropriate diameter can differ depending on whether the focus is momentum (pressure drop) or heat transfer, because the relevant perimeter in each phenomenon may not be the same. Given Data / Assumptions: Non-circular cross section (e.g., rectangular duct, annulus). For pressure drop, the wetted perimeter governs shear. For heat transfer, the heated perimeter governs convection to/from walls. In many practical exchangers, the heated perimeter ≤ wetted perimeter. Concept / Approach: The generic hydraulic diameter Dh = 4 * Area / Perimeter. For pressure drop, Perimeter = total wetted perimeter. For heat transfer, Perimeter = heated perimeter only. If heated perimeter is smaller than total wetted perimeter, then 4A / (heated perimeter) is larger than 4A / (wetted perimeter). Therefore, the equivalent diameter for pressure drop tends to be smaller than that for heat transfer. Step-by-Step Solution: Step 1: Write Dh,pressure = 4A / P_wetted.Step 2: Write Dh,heat = 4A / P_heated.Step 3: In typical cases, P_heated < P_wetted, so Dh,heat > Dh,pressure.Step 4: Conclude the relation: pressure-drop equivalent diameter is smaller than that for heat-transfer correlations. Verification / Alternative check: Annular passages with only the inner wall heated demonstrate Dh,heat computed with the inner perimeter, which is smaller than the sum of inner + outer perimeters used for pressure drop. This yields Dh,heat > Dh,pressure consistently. Why Other Options Are Wrong: Greater than: Reverses the typical inequality.Equal to: Only true for special cases where heated and wetted perimeters coincide.Not related to: They are related via area and perimeters.Variable but typically larger: Typical tendency is the opposite for pressure-drop Dh. Common Pitfalls: Blindly using a single Dh for both momentum and heat transfer without checking which perimeter applies can lead to errors in predicted friction factor or Nusselt number. Final Answer: smaller than

Question 19

Why are most industrial storage vessels and tanks designed as right circular cylinders?

Accepted Answer

Introduction / Context: Tanks and pressure vessels in chemical plants store liquids and gases and must withstand internal pressure and external loads. Shape selection affects stress distribution, fabrication complexity, and cost. Given Data / Assumptions: Thin-wall pressure vessel theory applies for many tanks. Fabrication uses rolling, welding, and standard head geometries. Cost and safety drive geometry choices. Concept / Approach: A cylinder distributes hoop stress uniformly under internal pressure and avoids stress concentrations typical of corners. Cylinders are also straightforward to roll from plate and to weld with standard heads (ellipsoidal, torispherical), enabling efficient, repeatable fabrication. Step-by-Step Solution: Step 1: Relate shape to stress: cylinders minimize peak stress vs prismatic shapes.Step 2: Relate shape to construction: rolling plates into shells and welding longitudinal and circumferential seams is standard practice.Step 3: Recognize cost and safety benefits: predictable stress distribution reduces required thickness for a given pressure, and standardization lowers fabrication cost.Step 4: Conclude both strength and fabrication advantages are decisive. Verification / Alternative check: Compare with rectangular tanks: corner regions would need heavy reinforcement; fabrication becomes complex and expensive for pressure service. Why Other Options Are Wrong: Only strength or only fabrication misses the full rationale.Neither: contradicts well-established design practice.Less floor area: not a primary governing reason. Common Pitfalls: Assuming spheres are always better: spheres minimize surface area but are far more difficult and costly to fabricate at large sizes; cylinders are the practical compromise. Final Answer: Both (a) and (b).

Question 20

In chemical process equipment, safety valves are primarily installed to protect against excessive __________ during operation.

Accepted Answer

Introduction / Context: Safety valves (or pressure relief valves) are last-line protective devices that automatically open to prevent equipment damage when internal conditions exceed safe limits. They are mandated by codes and standards for pressure systems. Given Data / Assumptions: Typical pressurized vessels, boilers, reactors, and piping systems. Primary hazard is overpressure due to upset, fire, blockage, or reaction runaway. Valve set pressure chosen below the maximum allowable working pressure. Concept / Approach: The safety valve senses pressure at its inlet. When pressure reaches the set point, it opens to discharge fluid, limiting the pressure rise. Temperature, turbulence, noise, and vibration are not the direct control variables of a spring-loaded safety valve; pressure is. Step-by-Step Solution: Step 1: Identify the control variable: safety valves actuate on pressure.Step 2: Recognize the purpose: prevent rupture by relieving excess pressure.Step 3: Select the option that explicitly mentions pressure. Verification / Alternative check: Design standards specify sizing and set points based on credible overpressure scenarios, confirming pressure is the parameter of interest. Why Other Options Are Wrong: Temperature: Indirect contributor; not the valve’s actuation input.Turbulence/noise/vibration: Possible symptoms but not protected directly by safety valves. Common Pitfalls: Confusing rupture disks and control valves with safety valves; only relief devices are certified for overpressure protection. Final Answer: pressure or pressure fluctuation

Process Equipment and Plant Design Questions